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 A344462 Number of pairs (d1,d2) of divisors of n such that d1 < d2 and d2^2 + d1*(n - d2)^2 = n^2. 0
 0, 1, 1, 2, 1, 3, 1, 3, 2, 3, 1, 7, 1, 3, 3, 4, 1, 7, 1, 5, 3, 3, 1, 9, 2, 3, 3, 5, 1, 9, 1, 5, 3, 3, 3, 10, 1, 3, 3, 7, 1, 9, 1, 5, 5, 3, 1, 11, 2, 5, 3, 5, 1, 9, 3, 7, 3, 3, 1, 13, 1, 3, 5, 6, 3, 9, 1, 5, 3, 7, 1, 13, 1, 3, 5, 5, 3, 9, 1, 9, 4, 3, 1, 13, 3, 3, 3, 7, 1, 13 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS If p is prime, then a(p) = 1. The only pair of divisors (d1,d2) of p such that d1 < d2 is (1,p), and since p^2 + 1*(p - p)^2 = p^2, (1,p) is the only solution. LINKS FORMULA a(n) = Sum_{d1|n, d2|n, d1

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Last modified May 26 04:26 EDT 2022. Contains 354074 sequences. (Running on oeis4.)